Question

Help for circle geometry question?!

Line OB is a tangent to the circle whose equation is `(x-5)^2-y^2=9`. A is the centre of the circle. Find the general equation of line OB.

Answer 1

`color(blue)(y=3/4x " and " y=-3/4x)`

Explanation:

I am assuming the equation of the circle should be:

`(x-5)^2+y^2=9`

`(x-5)^2-y^2=9` is not an equation of a circle.

If the line OB is tangent to the given circle it will have the form:

`y=mx` Since it passes through the origin.

Substituting this in the equation of the circle:

`(x-5)^2+(mx)^2-9=0`

Expanding:

`x^2-10x+25+m^2x^2-9`

Simplifying:

`x^2+m^2x^2-10x+16=0`

Arrange into the form `ax^2+bx+c`

`(1+m^2)x^2-10x+16=0`

Because we are just touching the circle at one point we need the roots of this quadratic to be repeated i.e. discriminant = zero.

Discriminant is:

`b^2-4ac`

`:.`

`(-10)^2-(4(1+m^2)(16))=0`

`100-(64+64m^2)=0`

`64m^2=36`

`m^2=36/64=9/16=>m=+-3/4`

So our equations are:

`y=3/4x and y=-3/4x`

There will always be two tangent lines that pass through a point outside the circle.

The general form of the tangent equation passing through the origin would be:

`y=mx and y=-mx`